Water flows through a horizontal tube as shown in the figure. If the difference of heights of water column in the vertical tubes is h = 0.02 m, and the area of cross-section at A and B are $4 \times 10^{- 4}$ m² and $2 \times 10^{- 4}$ m², respectively, then the rate of flow of water across any section is

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1.46×10^-4m³/s

1.70×10^-4m³/s

1.60×10^-4m³/s

1.30×10^-4m³/s

answer is B.

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Detailed Solution

$\begin{array}{l} v_{A} a_{A} = v_{B} \times a_{B} \Rightarrow v_{A} \times 4 = v_{B} \times 2 \\ \Rightarrow v_{B} = 2 v_{A} \dots \dots . (i) \\ Again, \frac{1}{2} {ρv}_{A}^{2} + {ρgh}_{A} + P_{A} = \frac{1}{2} {ρv}_{B}^{2} + {ρgh}_{B} + P_{B} \\ \Rightarrow \frac{1}{2} {ρv}_{A}^{2} + P_{A} = \frac{1}{2} {ρv}_{B}^{2} + P_{B} (as h_{A} = h_{B}) \\ \Rightarrow P_{A} - P_{B} = \frac{1}{2} ρ ({v_{B}}^{2} - {v_{A}}^{2}) = \frac{1}{2} \times 1000 \times (4 {v_{A}}^{2} - {v_{A}}^{2}) \\ \Rightarrow 1000 \times 10 \times 0.02 = \frac{1}{2} \times 1000 \times 3 {v_{A}}^{2} \end{array}$
(P_A-P_B=2 cm of water column )
$∴ v_{A} = \sqrt{\frac{4}{30}} = 0.3651 m / s = 36.51 cm / s$
So the rate of flow = v_Aa_A
36.51 x 4 = 146 cm³/s